Transcript for:
Understanding Motion in a Plane

hi friends you all know what is motion in a straight line but what if a body is not moving in a straight line if it goes in a zigzag manner like this then what type of motion it is that's right it is a motion in two Dimensions or what we call in physics as Motion in a plane now how do you describe the position Vector displacement Vector velocity or acceleration vectors for motion in a plane don't worry if all these words sound difficult to you I'm going to make these Concepts really easy in this video I'm going to explain these concepts with very simple examples so keep your pen and paper and let's get started but before we begin I just want to let you know that we have full courses on our website manocha academy.com and our Android App manocha Academy do check out the courses links are given below all right now let's look at Motion in a plane we have studied Motion in a straight line in the previous chapter for example if a car is moving along the x-axis like this then we can say the motion is in a straight line or let's say the car is moving along the Y AIS like this then also we can say motion is in a straight line in fact you can turn this x-axis over here and measure it in terms of X now for example if the car is moving like this in this XY plane what will you say is this Motion in a straight line or Motion in this plane do I really need the X and Y axis here no because you can see the motion is in a straight line and if I want to describe it in terms of X all I need to do is rotate this coordinate axis so I can make the x- axis aligned like this and then again I can describe the motion in ter terms of X so this is also Motion in a straight line but in this video we are going to look at Motion in a plane that requires X and Y AIS so let's take a look now let's say our car is moving like this along a curved path something like this is this Motion in a straight line obviously not because it's definitely not in a straight line so we need both X and Y axis to describe the motion of this car so we say the motion is in a plane plane means this two dimensional plane this flat surface here so what you're seeing this flat surface of the board we call it a plane we are right now in the two-dimensional plane and we need both X and Y AIS so we can say X and Y coordinates if we want to describe the motion of this car we cannot describe describe it using one dimension so this is called Motion in a plane and remember to think in 2D in terms of X and Y let's say our car was moving in this plane and after some time it is here at this position P let's say the coordinates of P according to this XY axis is 4A 3 so we have this XY coordinates here the cartisian coordinates for this plane and O there is the origin so I can describe the position of this car with respect to the origin right so we can draw a position Vector here from o to P like this and this is called the position Vector position Vector is usually denoted as r and since it's a vector we put the arrow symbol on it now of course the car is a large body so I've taken P to be the center of the car but you can think of a point object usually in this chapter we talk about a particle or a point object not a large size body but you can imagine here we have taken the center of this car okay so this is the position of this car and this is the position Vector op so how will you write the position Vector remember we have done the orthogonal unit vectors so unit Vector along the x-axis is called I and unit Vector along the y- AIS is called JC so the position Vector of this car at this position is going to be what since it is at position 4 comma 3 we will write the position Vector as 4 I + 3 jcap this is the position Vector at this position with respect to this coordinate system so in general how can we write the position Vector if an object is at X comma y we can say if anybody is at a position X comma y we can say the position Vector will be written as R cap = x i so x * the unit Vector I + y * the unit Vector J and this R Vector is called the position Vector here it is basically the vector op and we symbolize it as R now let's talk about the displacement Vector let's say our car moves from this position to this position so let's say it was at position 2A 3 here here the coordinates 2A 3 and it has moved to the coordinates 5A 5 with respect to this x and y axis now in displacement remember we don't care what path it took whether the car went like this or it went like this or it went straight it doesn't matter because displacement is always about the shortest distance from the initial to the final position so that is where we draw a straight line so the actual path doesn't matter so I'm going to erase this path here and we know our displacement is basically in a straight line from this initial to the final position now at the initial position remember we can describe the position Vector of the initial position so let's mark that so this will be the position Vector R1 for initial position and similarly you can draw another position Vector for the final position here and that's going to be R2 and you know that R1 is going to be 2 I + 3j R2 is going to be 5 I + 5j right we are interested in the displacement Vector which means from here to here this is the displacement right and we denote it as delt R the change in position so how do we describe delt R remember we have done vector addition so how will we get this R2 we know is basically r1+ delt R using the vector law of addition or I can say delt R the displacement Vector is R2 - R1 just like on the x-axis if you want to find the displacement from X1 to X2 what will you do final minus initial X2 - X1 same thing in the vector world you just have to subtract the final position minus the initial position the only fun is that these are vectors so our displacement Vector is going to be R2 minus R1 so can you guys find this Vector absolutely because we know our position vectors R1 and R2 so here we know R1 the position Vector R1 is going to be 2 2 I + 3 J so let's put that down in terms of the orthogonal unit vectors and the vector R2 position Vector R2 is going to be 5 I + 5j very simple so put that down as well and to get the displacement Vector you just have to subtract this so basically Delta R is going to be the displacement Vector Delta R is going to be R2 - R1 maybe I should have written R2 first and then R1 below that right but anyways you guys can do the subtraction so 5 i - 2 I should be 3 I and 5j - 3j is going to be 2 J there you go we have found the displacement Vector oh I forgot to put the arrow here Delta r or sometimes you'll see in the books it's written in bold that means it's a vector so this is the displacement Vector is 3 I + 2 J we have just found it simply using this formula right and again we can generalize this so this was for our simple example but let's say the car moves from X1 y1 to X2 Y2 what will be your displacement Vector very simple so in that case we can say displacement Vector delt R will be X2 - X1 time I the unit vector + Y2 - y1 times the unit Vector J so basically displacement Vector is what it is the change in displacement along the x- axis so Delta x i plus the change in displacement Y2 - y1 so Delta y * J there you go so this is the general expression for our displacement Vector so I hope with this example the concept of this displacement Vector here is clear to you when the body is moving from let's say position one to position two now let's look at average velocity so let's say the car moved from the position 2A 3 to 5A 5 and it took 2 seconds so we have been given extra information that the time is 2 seconds and we've been asked to find the average velocity over this motion so how will you do that even if you don't know Vector maths we can see that the car is having a displacement along the x-axis see it is moving along the x-axis and along the y axis that is why it has reached here right because it had a movement along the X and along the Y so we can maybe find the velocity along the x-axis and Y AIS and we know that they are perpendicular to each other right X and y- axxis are perpendicular so they are independent of each other so you can use that method first find the velocity along X and along y so what is the displacement along the x-axis you can see the car is moving from here to here so from x = 2 to x = 5 so we can see clearly the displacement or the change in X is 5 - 2 so Delta X is going to be three here and what is the change in y value the car goes from 3 to five see the y coordinate is three here it's five so the change in y Delta Y is going to be 5 - 3 which is 2 okay so what is the velocity along the x-axis velocity we know is change of displacement by time right rate of change of displacement so all you have to do is velocity along this x- axis will be Delta X divided by T or delta T change in time so that's going to be 3 ided 2 and let's say this is all in SI unit so it's 3 m by 2 seconds so 3x 2 1.5 m/s so simple we have found the velocity along this straight line what is the velocity along the y- axis again do the same thing the velocity along the y- AIS is going to be Delta y / delta T and if you put in the numbers what do you get 2 / 2 1 m/s so we have found the velocity along the X and Y AIS so I can combine these two using our orthogonal unit vectors and I can give it a vector notation now so we can say the average velocity V is going to be vxi 1.5 5 * I the unit Vector we are giving it a direction because the job of the unit Vector is to give it a direction this is the direction along the x-axis plus 1 * the unit Vector J so I'll just write J so there you go we have got our average velocity so let me write AV over here for average velocity so this is one method a two-dimensional motion you can break it down into motion along the X and motion along the Y break down get the values and then put your I inj but now you guys are starting to become expert at vectors so we can use the vector notation Also let's take a look at that so if I say what will be the average velocity using Vector notation so we can write it as average velocity is going to be delt R so average velocity is going to be delt R by delta T and remember we had found this delt R the displacement Vector what was delt R so we had found this displacement Vector delt R which was R1 - R2 if you do that it's basically 3 I + 2 J so you have to do 5 - 2 so that's going to be 3 I + 5 - 3 2 J so you can substitute that here right and directly get your values so if you put that here delt R is 3 I + 2 J so we are directly using Vector notation time is a scalar so divided by 2 seconds that is given to us so if you do the division see you'll get the answer directly 1.5 * the I Vector plus 1 * the J Vector this will be our average velocity and this is the exact same answer we got when we did with our component method where we were finding the velocity along X and Y and combining it so you can directly start using this Vector notation and get the answer so we can generalize this example so what will be the average velocity if a body moves from let's say X1 y1 to X2 Y2 in a time interval t What will be the general formula we can say average velocity is delt R by delta T and what is delt R the change in the X displacement Delta x * I plus change in y displacement * J so I can put Delta x i plus Delta y j cap divided by delta T now if you divide these individual terms Delta X by delta T Delta y by delta T what will you get Delta X by delta T is the velocity along the X Direction so we can put that as V X this is the scalar value right this is the magnitude of the Velocity along x * I which is the direction plus Delta y by delta T is nothing but velocity along y * G so don't get scared of vectors start getting familiar with this notation it's not difficult if you just start seeing everything in terms of i j right because they're independent of each other so you can see it is finally working out to the concept we know that we can add up the velocity along the x-axis times the I Vector plus the velocity along y AIS times the J Vector to get our overall average velocity we have discussed about average velocity now let's look at instantaneous velocity the velocity at a particular instant so how will you work that out let's say once again our car is moving from here to here so we can say our displacement from this this position to this position if we draw the vector that is our displacement Vector delt R and once again let's say the time interval is delta T so we know average velocity is going to be the average velocity is delt R the change in displacement divided by the change in time delt R by delta T but if we want instantaneous velocity that means at a particular instant then we need to make this time interval very very very small like 00001 second very small right so in the language of calculus we write it as when limit delta T tends to zero that means this time interval is very small then we say it becomes the instantaneous velocity and we denote it as Dr by DT so you're doing a differentiation of the r Vector here now let's understand what that means if we look at the motion of this car so let's say the car was moving like this so let's say this was the motion of the car okay from this position to this position if you want to find the instantaneous velocity that means let's say you want to find it from at this instant so you will take a very very small time interval so let's say we zoom in and we say that this is the initial position and let's say this is the final position okay you have zoomed into this uh displacement of the car and we want to know what is the speed like what is the speed that you see in the speedometer of this car at this instant so physics tells us that you need to take a very small time interval and you need to divide the displacement by that time interval so when you do this what will your delt R Vector look like the delt R is going to be from here to here right this is the displacement so this is my delt R vector and we know time is a scalar so instantaneous velocity will have the same direction as delt R the change in displacement so what will be the direction of the instantaneous velocity for that we can draw this we can say this is is the direction here and what is this exactly this is a tangent to the curve at that point so this is a very important concept because you are drawing a line when the time interval is very very small you're drawing a line which just touches the curve so this is the direction of your instantaneous velocity and you can find the value using Dr by DT so similarly what will be the instantaneous velocities Direction here again you draw tangent what will be the instantaneous velocity direction here you draw a tangent at this point right and we can also use the slope concept here because we know it is Dr by DT so velocity is the slope of the displacement time graph right you guys know that and this is the direction of the Velocity tangent at that point and if you want to know the angle involved here what is is the formula you guys know that tan Theta the angle made by this tangent tan Theta will be perpendicular by base so here we are talking about velocity so you can think this is the velocity VY along the Y AIS and the velocity VX along the x-axis and this angle I'm looking at tan Theta is going to be VY by VX right because using trigonometry perpendicular divided by base will be your value of tan Theta here so remember direction of instantaneous velocity at any point in the path is the tangent at that point as we have discussed here and once again you can apply your vector maths here so when we expressed it as Dr by DT if you substitute the r Vector here so if you substitute r as x i + y j so what do we get here if you do differentiation of that basically you'll get DX by DT * I plus Dy by DT * J so remember this trick always in motion in a plane you can always split it up into the motion along x- axis and y axis because they are orthogonal to each other perpendicular to each other so they are independent right that's why you'll see in the differentiation we get DX by DT right the uh displacement along the xaxis or you can think of the instantaneous velocity along x-axis times the I Vector plus the instantaneous velocity along y- axis Dy by DT time the J Vector so that is basically your instantaneous velocity we have looked at displacement and velocity now let's look at acceleration so first we'll talk about average acceleration once again let's say the car is moving from here to here and at this point it's velocity is let's say I + J so we can say the initial velocity V1 is I + J here and the final velocity of the car here V2 is 4 I + 3j note that these are the velocities these are not the position position of the car so this is the velocity Vector here I + J and this is the other velocity Vector here 4 I + 3j so we want to find the acceleration of the car and let's say the time it took is 3 seconds so the information given to us is the car changes its velocity from position one to position two in 3 seconds and your job is to calculate the average average acceleration so once again we'll apply the concepts that we know that acceleration is rate of change of velocity right so if you want to find acceleration we can say acceleration is going to be Delta V by delta T what is Delta V the change in the velocity Vector so once again you could break it down and find the change in X and change in y or you can apply your vector mathematics here so we can say change in velocity is simply going to be V2 minus V1 or you can treat it as V minus U final velocity minus initial velocity so what will Delta V be simple I just need to subtract the two vectors V2 minus V1 so these are all vectors right V1 and vs2 and we need to do [Music] V2 minus V1 so if you subtract what will you get 4 I + 3 J minus I + J so if you do the subtraction what do you get 3 I + 2 J do the subtraction 4 i - I is 3 I and there you have 2J so we are getting 3 I + 2J here this is the change in velocity and for acceleration we need to divide it by time so let's do that so acceleration is therefore going to be be Delta V by delta T so basically I do 3 I + 2 J and let's divide the component each component with the time three so the final answer is going to be I + 2x3 [Music] J so this is our average acceleration over the time 3 seconds so simple just apply the vector maths and there you got it so this was for this example and once again we can generalize it right for any velocity change from V1 to vs2 what will the formula be acceleration is going to be the change in velocity so Delta V by delta T and here we can say we can break down the change in velocity as the change along the X Direction so Delta VX * I plus the change in the y direction so Delta V in the y direction time J / delta T and Delta VX by delta T is nothing but the change in the velocity along the X Direction so it was basically 4 I minus I right so this we can say is the acceleration over here the acceleration along the X Direction ax I and this is the acceleration along the y direction a y j so there you get so this is the general notation that the net acceleration the average acceleration is the acceleration along the x-axis so times I plus the average acceleration along the Y AIS time J so this is how you work out your average acceleration when the body changes its speed from V1 to vs2 again remember these are all vectors acceleration is a vector quantity so finally you get a vector if you want to find its magnitude you can do that square root of ax s + a y squ right and you can find its direction by doing tan Theta is a y by ax so all those Vector fundas that we have learned will apply here for velocity acceleration displacement and this is how you express them we looked at average acceleration now let's talk about instantaneous acceleration instantaneous acceleration means acceleration at a particular instant so how do we work that out so remember average acceleration let's say a body was moving with a velocity V1 and this car uh achieved a velocity of vs2 let's say in time T seconds so what is the average acceleration we looked at that average acceleration will be the change in velocity by change in time because acceleration is the rate of change of velocity now if you want the instantaneous acceleration that means while the car is in motion right so let's say this is the motion of the car at a particular instant you want that acceleration so remember what do we need to do we need to make the time interval very very small just like instantaneous velocity like 0.001 second so in a very small time interval and in calculus we write it as limit delt T tending to Z so in that limit delta T tending to zero what will this formula turn out to be it's basically going to be DV by DT because when the time interval is very small this becomes the differentiation of velocity or we say the derivative of velocity with respect to time and this is what we call as instantaneous acceleration so basically it is telling us the acceleration at any particular instant is your instantaneous acceleration and again remember this is a two-dimensional motion right it's in the car is moving along the X and along the Y so once again we can break it down like this so we can write DDT and if we express velocity in terms terms of VX I plus VY J so you can do the differentiation here in terms of X and Y so you'll get DDT of VX * I plus DDT of v y time J and this will give you your instantaneous acceleration so you can always break it down into the horizontal X component and vertical y component like this and remember acceleration is a vector so we have given its direction in terms of the orthogonal unit vectors here I and J so this is the concept of instantaneous acceleration the rate of change of velocity in a very very short interval of time we looked at instantaneous acceleration because acceleration could be changing as well just just like velocity right a body need not be moving at a uniform or constant acceleration but now we look at that special case where the body is moving at a constant acceleration how can we calculate the velocity here so let's look at this case let's say it's given to us the initial velocity which we denote by U of the car is 2 I + J and the car moves for 2 seconds and the acceleration of the car is given as 3 I + 4J now you may be thinking why this Vector notation because remember acceleration is a vector so it has a magnitude and a Direction so I and J vectors will help you define the direction it is 3 I + 4 J of course you can calculate the Direction with tan Theta uh equal to Y by x value right but note that the acceleration here is constant it is fixed so the acceleration that has been given to us is a constant value it does not change and our goal is to find the velocity of the car after 2 seconds so this is our final velocity V and that's what we need to find so how will you do this question let me teach you a very simple trick where you can break down your two dimensional motion into onedimensional motion how because we know that we can break it down into the horizontal motion and vertical motion because they're independent so along the x-axis and y- axis so let's break that down so if we want to find the final velocity we'll find the final velocity along the X Direction and Y Direction and just add up the two with Vector notation so let's do that so we can say along the x-axis what do we know we know that the initial velocity U is 2 I + J so the X component is two right and again we can take SI units let's say meter per second so along the X axis I know that my U is 2 m/s okay what is given to us time is given as 2 seconds so time is 2 seconds how much is the acceleration along the x- axis so here you can see it's given 3 I + 4 J that means the acceleration along the x axis so these are all X components so ax I can say is three since we are doing all along x- axis now we are not writing 3 I we just writing the values and let's say this is SI unit so 3 m/s square and we need to find V and this is very easy for you because you can use the v = u+ a the first equation of motion here because this is motion in a straight line along the x-axis and again we want to find basically VX so what will it be our v = u + a or you can write all these X over here time is a you know same for both so we don't have to write TX now just substitute and you'll get the values so 2 m/ second + 3 m/s squ times the time is given as 2 seconds so finally the final velocity along x- axis is 2 + 6 which is 8 m/s so x axis is done similarly find the velocity along y AIS same procedure so we are going to write y AIS the initial velocity along y- axis is how much see it's given plus J here right so it's basically 1 m/s time again same thing 2 seconds acceleration what is the acceleration here 4 along the Y AIS so 4 m/s square and we want to find v y the final velocity along y- AIS again use the same formula v = u + a so v y = uy + a y oops I wrote X over here by mistake this should be a y time T so just substitute the values very simple so this will be 1 + 4 * 2 so 1 + 8 is 9 m/s that is your VY so see for the body having constant acceleration we can easily work out the final velocity here by breaking down the motion into the X component and the Y component this is what we have done here basically and so the final answer is going to be what you just combine these two if you want to write it in Vector notation because you know velocity is a vector so the final velocity V is going to be VX I plus v y j because these were the values and we are giving it Direction with the orthogonal unit vectors I and J so that's it now if we substitute VX uh this is 8 I plus how much do I have there 9 J finished now this is a vector if you want to know what is the magnitude of the Velocity how will you find it Square < TK of 8 square + 9 squ right we can easily find the magnitude of the vector it is going to be square < TK of 8 s + 9 s so you guys can work it out m/s if you want to find the direction of the Velocity Vector that is also very simple what we do tan Theta because we measure the Direction with the x- axxis the slope it makes with the x- axis that's going to be y by X so 9 by 8 and then you can use the trigonometric tables or calculator to find out your angle Theta so see it is so simple the main trick here is we broke down the two dimensional motion into two onedimensional motion along X along y applied the equations and there you can see we easily have our answer here so let's generalize what we have learned let's say we want to find the velocity when the motion has constant acceleration a fixed acceleration and again we are talking about a two-dimensional motion so let's say the initial velocity U or you can write it as V1 is uxi + u j and the acceleration is axi + a y j so this acceleration is basically constant it has a fixed value and we want to find the final velocity after a time interval T so what will be the final velocity again we can say generalizing it is in the X and Y Direction so we can write it as VX I plus v y j and as we discussed how do you calculate this final velocity break it down right into the X component and into the Y component so we are basically doing resolution of the vectors we are breaking it down into the orthogonal components so what will be the VX value VX is going to be equal to ux that is the horizontal veloc initial velocity plus the horizontal acceleration multiplied by the time T so this will help you calculate the final velocity along the X direction that is this is your VX and if you want to calculate VY the final velocity along the Y Direction so VY is going to be the formula is the initial velocity along the y direction plus the acceleration along the y direction multiplied by the time T and so when we take these two quantities you just multip mped with the I and J you will get your velocity Vector so this is the general notation of what we have learned and again remember break down the two-dimensional motion into two onedimensional motions what are they the X and Y component that is why these orthogonal unit vectors or doing this resolution of vectors into these orthogonal components is very very useful and makes your calculations analysis really easy so we have seen how to calculate final velocity when acceleration is constant now let's take a look how we can calculate the displacement so let's say the car is moving in this two-dimensional motion and it starts off at a point 1A 1 so that is its initial position and let's say the initial velocity is 2 I + J and it is moving with a constant acceleration we can take the same value 3 I + 4 J and we need to find the displacement of the car after time 2 seconds so T is 2 seconds over here we need to find this final position of the car after 2 seconds so how will you do this again I'm going to teach you the simple trick that do not complicate it with two-dimensional motion you can just break it into two onedimensional motion just like we should make our life simple make these questions simple so which equation should we use here again use that trick of equations of motion first take a look what has been given and what you're asked so we have been given acceleration a Time T we know the initial velocity U right we also given the initial position but we need to find the displacement here so basically the unknown is the S so our unknown is the displacement of the body normally we not displacement with s but usually in Vector notation we displacement uh but usually in Vector notation displacement is denoted as or change of displacement as delt R so if we can find this delt R over here the change in displacement we can find the final position right so which equation of motion should we use so basically u a is known and we need to find Delta r or you can call it the displacement s over here so that's right we need to use the second equation of motion which is s = UT + half a² and once again we will not work in two dimensional motion we are going to break it into the x-axis and the Y AIS right motion along X and Y so let's go ahead and do that we'll start off with the xaxis over here so along the x-axis what is the initial velocity guys you can see U AIS 2 so let's put down initial velocity is 2 m/s time given to us is 2 seconds right and acceleration what is the acceleration along X acceleration is going to be three so we'll put that down here ax is 3 m/s squ and we need to find basically our displacement along X so I'm going to call it s with the subscript X over here so let's work it out so s along xaxis is going to be ux plus uxt + half a t² actually you can write the equation and then put all the X it's more convenient right so we are just copying down this equation and just putting the subscript X now just let's substitute the values so 2 * time is 2 here so 2 * 2 + half * 3 * 2 * 2 2 2 cancel so what are we getting 4 + 6 10 and the unit will be meters we are in SI units so displacement along X is 10 m along the horizontal X x-axis find the y axis also go ahead and try that yourself so what will it be y AIS is going to be same thing over here uy is 1 m/s time 2 seconds the acceleration along y AIS a y is 4 m/s square and we need to [Music] find the displacement along y once again use this so what is Sy y going to be again you can write s = to UT + half a² and put the Y subscript over here so go ahead and substitute the values what will it be 1 * 2 2 so this is going to be 2 + half * 4 * 2 2 so 4 * 4 will give you 16 by 2 is 8 so this will give you 8 we are getting 10 m interesting we are getting both SX and Sy y as 10 M over here so that is the displacement along the X and Y axis but the question was asked to find the final position let's say they want to know where is the car after this 2 seconds so how will you do that very simple this was your displacement along the x-axis so you know how much is Delta X and you know how much is Delta y so to get the final position you just need to look at your initial position so it will be basically 1 plus Delta X similarly 1 plus Delta y so so simple you just add these displacements because we know that this is 10 m this is also 10 m or 10 units here this is 1A 1 so the final position this position is therefore going to be 10 + 1 and 10 + 1 this is going to be 11a 11 so use this simple technique simplify your life break down the motion in a plane two dimensional motion again I'm repeating again and again break it down into X and Y components then you it becomes so easy you do not have to use any uh Vector stuff here you just break it down and there you're done if you want to express the displacement you know or you want to express the position in terms of the position Vector then of course we can write R is going to be 11 I + 11 J similarly if you wanted to express Delta R what is the displacement of the body we can say it is 10 and 10 so 10 I + 10 G see it's so simple now let's apply the concepts that we've learned to solve this question so the question says a body is moving Eastward with a velocity of 5 m/s so let's visualize this always draw a diagram in physics to understand the question so what does it say a body is moving towards the east this is the East direction right and it says with a velocity of 5 m/s so we can say the velocity is 5 m per second here in 10 seconds the velocity changes to 5 m/s northwards so after 10 seconds the velocity becomes like this 10 5 m/s towards the North and this happens in time of 5 Seconds sorry 10 seconds and we need to find the average acceleration of the body so how will you guys do that remember acceleration is the rate of change of velocity Delta V by delta T so you can see this is our initial velocity so we can mark it as U or you can use V1 whatever you like so U for initial velocity and this is V our final velocity is 5 m/s since I have just the values here I'll not use the vector notation for now so we are just going to write u and v here acceleration remember we have to calculate the average acceleration so average acceleration the formula is Delta V the change in the velocity divided by delta T So how will you find the Delta V guys remember velocity is a vector it has direction also so and we know the change in velocity is going to be final velocity minus initial velocity which is V minus U so we know that Delta V guys is going to be V minus U but these are vectors we have been only given the value but we've also been given the direction so let's put it into Vector notation very easy and once again you can use your orthogonal vectors so we can use our orthogonal unit vectors if you imagine the XY plane this is along X this is y this is the negative x- axis and negative Y axis so we know that unit Vector along X is i unit Vector along Y is J so my what will be my initial velocity 5 m towards the positive x-axis so this is 5 I actually right so this initial velocity is basically 5 I and the final velocity is 5G because it's in the y direction right so we can write this as 5 J - 5 I that is my change in velocity and time delta T So t or delta T you write is 10 seconds so just substitute it and you will get your average acceleration so average acceleration is going to be this Delta V Delta V by delta T so 5 J - 5 I / 10 seconds so you have to divide each individual component with 10 so this is 5X 10 J minus 5x 10 I this is basically half of J so 1X 2 J - 1X 2 I so there you go you have got your average acceleration vector this is your average oh not velocity average acceleration vector a but let's say you want to give the magnitude and the direction of acceleration here we have expressed it in Vector notation so this is also a correct answer but sometimes they tell you or you supposed to find the magnitude and Direction so how do we do that it's a vector how do we find the magnitude of vector very simple so what we can do here to find the magnitude of the acceleration I'm going to to do square root of the X component plus the Y component right now it's only magnitude so it will be square of half square + - half square and the square root of that so square root of what is half Square 1/4 + 1/4 the negative sign goes away so 1/4 + 1/4 is half so we are getting Square OT of half which is nothing but 1 by < tk2 or if you want to rationalize the denominator you can multiply by < tk2 and < tk2 so it's basically going to become if you multiply by ro < tk2 and by < tk2 we are basically getting < tk2 by 2 m/s squar so again I'm taking all SI units here this is the magnitude of the acceleration now what is the direction of the acceleration So Tan Theta is going to be y by X so Y is - half X is half here so let's substitute tan Theta is a a y by ax which is which is - half / half so that's going to be min-1 but rather than leaving it like this let's see if we can work out the direction when is tan Theta equal to minus1 we know tan Theta is + 1 when Theta is 45° and it's minus1 when Theta is 100 35° right so it is making an angle with the x-axis of the acceleration vector is going to be something like this your acceleration vector is making an angle of 135° with the xaxis so now if you apply your geography of north south east west can you guys tell me what is the direction of acceleration that's right it's along the Northwest Direction so our acceleration is along the so we can say the direction of acceleration direction of a is Northwest n w so see it becomes so simple if you use this powerful orthogonal unit vectors you can easily do the question we don't need to do some geometry here and calculate just go with Vector notation work it out and you can easily find the acceleration we found its uh magnitude here see magnitude is < tk2 by2 m/s squ that's the acceleration vector and here's the direction in the Northwest Direction now let's go ahead and try this question the position of a particle is given by the position Vector R which is 3 TI I + 2 t² J + 5 K where t T is in seconds so T is the time in seconds and the coefficients have proper units for the r to be in meters so basically everything is in SI units here we need to find the velocity VT and the magnitude and the direction of the Velocity at t equal to 3 seconds so you have to do two things here find the velocity Vector T in terms of T and the velocity at time t equal to 3 seconds so how do we do this let's apply the concepts we have learned so first thing we will write down is this position Vector R so R is given as 3 TI I see it's dependent on I because as the body is moving according to uh according to the time T the position Vector is changing so 3 TI I + 2 t² J + 5 K now how do you find the velocity velocity is rate of change of displacement or rate of change of the position Vector so very simple the velocity V is going to be Dr by DT why am I doing this because I'm finding the instantaneous velocity right we want to express it in terms of T that means the instantaneous velocity you need to recognize that they will not tell you always average or instantaneous so here we need to find the instantaneous velocity drdt because it is at any time T okay so what you'll do here you'll take a derivative of this differentiation so DDT and just substitute this R over here so we'll put in 3 TI plus I'm not putting all the I cap J cap K cap I'm feeling a bit lazy here so I'll just write 3 TI I plus but don't read it as TI right this is I is the unit Vector here + 2 t² J + 5 K okay so now now you can break down this differentiation it will become DDT of these individual quantities right so it's basically DDT of 3T which is take out t uh 3 3 is the constant so DDT of t i plus we have to do differentiation of this so 2 * DDT of t² J plus DDT of 5K so 5K is a constant so this will be simply zero so now go ahead and simplify this and you will get your velocity Vector very easily so velocity vector v is going to be DDT of T is 1 so 3 I plus what do we have your DDT of t² what is that going to be by differentiation you'll get 2T 2 T multiplied by 2 4 T So 4T of G and here we have zero so we don't have to write 0 K so this is our velocity vector v or you can write it as VT C why VT because it is in terms of T that means the velocity is constantly changing this is the instantaneous velocity at every time T the velocity will be different so this is our final answer for our expression VT so we have done the first part now how do you do the second part you have to calculate the magnitude and direction of the Velocity at time T = 3 seconds so go ahead and use this expression and see if you can find it out so this is what we had calculated in the first part this is our expression of the Velocity the instantaneous velocity and now we need to find it at tal to 3 seconds so very simple just substitute this 3 second into the expression here so what will we get the velocity at 3 seconds that means we can write V of 3 is going to be 3 I + 4 * 3 J so very simple that's going to be 3 I + 12 J so this is our velocity Vector at 3 seconds but the question is asking so this is our velocity Vector at 3 seconds but the question is asking us find the magnitude and direction of this velocity Vector so how do you do that simple it's a vector you know how to find magnitude and Direction the magnitude is going to be square root of the X and Y component here squared right so square otk of 3 square + 12 squ so you will have square otk of 3 squ + 12 square over here which is square < TK of 9 +44 that is basically square root of 1550 3 so this is the magnitude of velocity Square OT of 153 m/ second did you know 153 is a very interesting number it's called Armstrong number why because if you do 1 Cube + 5 Cub + 3 CU you're going to get 153 you guys can try that out anyway so here we have got the magnitude of the Velocity here and what will be the direction of the Velocity very simple because this is our velocity this is the X component and this is the Y component so we can find the angle tan Theta in terms of vertical component VY by VX so just plug that in here so that will be tan Theta is 12 / 3 so we can say tan Theta = 4 or Theta is tan inverse 4 so you can use the trigonometric tables to find the answer so here we have these final answers the magnitude of the velocity and we have the angle of the Velocity see so simple we are using the vector Concepts that we had learned if you haven't seen my videos do watch that and we have discussed all these in detail over there here's the same question but this time you guys need to find the acceleration at of the particle the position Vector is given to us and you need to work out the acceleration why don't you try this question yourself and do let me know your answer by putting it in the comments below I'm looking forward to reading your answers so try this homework question and put your answers in the comments below